Question: $2x^3+x^2y+y^2=4$ Find the value of $\dfrac{dy}{dx}$ at the point $(-1,2)$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $\dfrac{2}{5}$ (Choice C) C $-\dfrac{2}{5}$ (Choice D) D $-2$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $2x^3+x^2y+y^2=4$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} 2x^3+x^2y+y^2&=4 \\\\ \dfrac{d}{dx}(2x^3+x^2y+y^2)&=\dfrac{d}{dx}(4) \\\\ \dfrac{d}{dx}(2x^3)+\dfrac{d}{dx}(x^2y)+\dfrac{d}{dx}(y^2)&=0 \\\\ 6x^2+\left(2x\cdot y+x^2\cdot\dfrac{dy}{dx}\right)+2y\cdot\dfrac{dy}{dx}&=0 \\\\ 6x^2+2xy+x^2\cdot\dfrac{dy}{dx}+2y\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 6x^2+2xy+x^2\cdot\dfrac{dy}{dx}+2y\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(x^2+2y)&=-(6x^2+2xy) \\\\ \dfrac{dy}{dx}&=-\dfrac{6x^2+2xy}{x^2+2y} \end{aligned}$ Now we can plug the point $(-1,2)$ into the expression for $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=-\dfrac{6x^2+2xy}{x^2+2y} \\\\ &=-\dfrac{6(-1)^2+2(-1)(2)}{(-1)^2+2(2)} \gray{x=-1,\,\,y=2} \\\\ &=-\dfrac{2}{5} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at the point $(-1,2)$ is $-\dfrac{2}{5}$.